why doesn't the north star move?

i need the answer asap and i will give brainliest​

Answer:

Refraction. A light ray traveling through some plastic has a frequency of 5.5 x 1014 Hz. It is ... making an angle of 32° with the normal to the interface; it passes through the ... [5 points] (a) What is the refractive index of the glass? ... What you do know is the wavelength of the light in glass, 321 nm, and you know the frequency.

Explanation:

The refractive index of Acrylate is determined as: ratio of sine of incident angle to  sine of refractive angle.

The relationship between the angles of incidence and refraction for light or other waves flowing through a border between two different isotropic media, such as water, glass, or air, is described by Snell's law, a formula.

The amount a light ray bends as it travels through different media is measured by the refractive index.

Given that: The refractive index for air is 1.00.

Let the angle of incidence from medium air to Acrylate  is = i

the angle of refraction  from medium air to Acrylate  is = r.

Let the refractive index  for Acrylate  = n.

Then according to Snell's law:

1×sini = n× sinr

n = sini/sinr.

Hence,  the refractive index of Acrylate is ratio of sine of incident angle to  sine of refractive angle.

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The ball passes through half the maximum height on the way up and down at approximately 1.28 seconds.

To determine the maximum height reached by the baseball and the times at which it passes through half the maximum height on the way up and down, we can use the equations of motion for vertical motion.

Maximum height (hmax):

The initial velocity (u) is 12.5 m/s and the acceleration due to gravity (g) is -9.8 m/s^2 (negative because it acts downward). The maximum height reached by the ball can be calculated using the equation:

hmax = (u^2) / (2g)

Substituting the given values:

hmax = (12.5^2) / (2 * 9.8) ≈ 8.04 meters

Therefore, the maximum height reached by the ball is approximately 8.04 meters.

Time at half maximum height (t1/2, up and t1/2, down):

The time taken for the ball to reach half the maximum height on the way up and on the way down will be the same. We can calculate this time using the equation:

t1/2 = (u - v) / g

Where v is the final velocity, which is 0 m/s at the highest point of the trajectory.

t1/2 = (12.5 - 0) / 9.8 ≈ 1.28 seconds

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The terminal velocity of the sphere is approximately 22.68 mph.

The terminal velocity of a sphere is the constant speed at which the gravitational force pulling the sphere down is balanced by the drag force pushing the sphere up. The drag force is proportional to the velocity of the sphere, and can be calculated using the following formula:

Fd = (1/2) * rho * Cd * A * v²

where Fd is the drag force, rho is the density of the fluid (air in this case), Cd is the drag coefficient (which depends on the shape of the object), A is the cross-sectional area of the object perpendicular to the direction of motion, and v is the velocity of the object.

The gravitational force pulling the sphere down is given by:

Fg = m * g

where m is the mass of the sphere and g is the acceleration due to gravity.

At terminal velocity, the drag force is equal in magnitude to the gravitational force, so:

Fd = Fg

Substituting the expressions for Fd and Fg and solving for v, we get:

v = √((2 * m * g) / (rho * Cd * A))

where A = pi * r² is the cross-sectional area of the sphere, and r is the radius of the sphere.

Plugging in the given values, we get:

v = sqrt((2 * 0.6 * 9.81) / (1.2 * 0.47 * pi * 0.3²)) ≈ 10.13 m/s

To convert this to mph, we multiply by 2.23694:

v ≈ 22.68 mph

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If a jogger runs a 10 kilometer race in 60 minutes, her average speed is 10km/hr. Details about average speed can be found below.

Average speed can be calculated by dividing the distance moved by a body by the time taken. That is;

Average speed = Distance/Time

According to this question, a jogger runs a 10 kilometer race in 60 minutes. The average speed is calculated as follows:

Average speed = 10km/1hr

Average speed = 10km/hr.

Therefore, if a jogger runs a 10 kilometer race in 60 minutes, her average speed is 10km/hr.

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Answer:

m=image distance÷object distance

i) The initial deceleration of the car is -1.6 m/s² and  the time taken is 5 seconds

ii) The final deceleration is -1 m/s²

iii) The total dispalcement = 1016 m

The initial deceleration of the car is given by the formula below:

v² = u² + 2as

where;

Solving for a;

12² = 20² + 2 * a * 80

a = -1.6 m/s²

Time taken, t = v - u / a

t = 12 - 20 / (-1.6)

t = 5 seconds

Final deceleration:

a = v - u / t

a = 0 - 12 / 12

a = -1 m/s²

iii) Displacement at constant velocity = 12 * 1 * 60

Displacement at constant velocity = 720 m

Final displacement, s = ut + 0.5at²

s = 12 * 12 + 0.5 * 1 * 12²

s = 216 m

Total dispalcement = 80 + 720 + 216

Total dispalcement = 1016 m

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The minimum coefficient of static friction with the floor is 0.3846.

To find the minimum coefficient of static friction with the floor, we need to consider the forces acting on the couch. In this case, the force of gravity is pulling the couch downward with a magnitude of mg, where m is the mass of the couch (40 kg) and g is the acceleration due to gravity (9.8 m/s²).

Since the couch does not move, the force of static friction between the couch and the floor must be equal in magnitude but opposite in direction to the horizontal pushing force of 150 N.

Therefore, we have the equation F_friction = F_push, where F_friction is the force of static friction.

The force of static friction can be calculated using the formula F_friction = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force.

Since the couch is on a level floor and is not accelerating vertically, the normal force N is equal in magnitude but opposite in direction to the force of gravity, which is mg.

Substituting the values into the equation, we have μs * mg = 150 N.
Solving for μs, we get μs = 150 N / (mg).
Substituting the given values, we have μ_s = 150 N / (40 kg * 9.8 m/s²).
Simplifying, we find that μs = 0.3846.

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The current density of a 7.0 mm diameter copper wire that carries a current of 7.0 μA is 18.19 × 10⁻³ A/m².

The quantity of electric modern-day journeying in step with the unit move-phase region is known as current density and is expressed in amperes per rectangular meter. The more present-day a conductor, the higher could be the modern-day density. current density is the quantity of electrical modern flowing in step with the unit move-sectional region of a material.

subsequently, the SI unit of current density needs to be Ampere/meter2. Ohm's law relates the current flowing via a conductor to the voltage V and resistance R; this is, V = IR. An alternative assertion of Ohm's law is I = V/R.

current I = 0.7 μA

r = 3.5mm

J = I/A = I/πr³

J = 0.7 × 10⁻⁶ / 3.14 × (3.5 × 10⁻³)²

J = 18.19 × 10⁻³ A/m²

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Answer:

the substitute bride is a good girl who knows

1. The time taken for the car to stop is 5 s

2. The distance travelled during the reaction time is 50 m

3. The distance travelled during the stopping time is 62.5 m

1. How do I determine the time taken for the car to stop?

The time taken for the car to stop can be obtained as follow:

a = (v - u) / t

-5 = (0 - 25) / t

-5 = -25 / t

Cross multiply

-5 × t = -25

Divide both sides by -5

t = -25 / -5

t = 5 s

Thus, the time taken to stop is 5 s

2. How do I determine the distance travelled during the reaction time?

The distance travelled during the reaction time can be obtained as follow:

Distance travelled = velocity × reaction time

Distance travelled = 25 × 2

Distance travelled = 50 m

3. How do I determine the distance travelled during the stopping time?

s = ut + ½at²

s = (25 × 5) + (½ × -5 × 5²)

s = 125 - 62.5

s = 62.5 m

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Gases have heavier molecules. Since all gases have the same average kinetic energy at the same temperature, lighter molecules move faster and heavier molecules move slower on average.

Answer:

Aerobic exercise became part of this movement in the 1970s.

Answer:

D.barko po

Explanation:

I trying my best

The displacement of the runner after four seconds is 78.4 m.

We know that in this case, we are dealing with a case of an object that has a motion under gravity. We are told that the road runner an fell straight down off a cliff. The fact that we have been told that the runner just fell down the cliff means that the initial velocity of the runner would have to be taken in this context as zero since the runner was dropped from a height as shown.

Acceleration of the runner (g) = 9.8 m/s^2

Initial velocity of the runner (u) = 0 m/s

Time take (t) = 4 seconds

We then have;

h = ut + 1/2gt^2

If we then know that the initial velocity of the person is zero, then we have;

h = 1/2gt^2

h = 0.5 * 9.8 * (4)^2

h = 78.4 m

The height is 78.4 m.

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Answer:

the war of currents is

Explanation:

The war of currents was a series of events surrounding the introduction of competing electric power transmission systems in the late 1880s and early 1890s.

Answer:

a series of events surrounding the introduction of competing electric power transmission systems in the late 1880s and early 1890s

Explanation:

Answer:

Blank 1: Gasses

Blank 2: More

Blank 3: Solids

Blank 4: Fluids

Blank 5: Liquid

Blank 6: Gas

Blank 7: Higher

Blank 8: Lower

Blank 9: Sun

Blank 10: Radiation

Blank 11: Conductors

P.S. order of answers does not matter between Blank 5 and 6.

I Think Its Real

A wide range of empirical evidence strongly favors the Big Bang event, which is now essentially universally accepted. Detailed measurements of the expansion rate of the universe place the Big Bang singularity at an estimated 13.787±0.020 billion years ago, which is considered the age of the universe.

Explanation:

15-letter words that start with e

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The speed of the roller coaster car as shown is  31 m/s

We know that the roller coaster is the kind of device that we can use to be able to show the conversion of kinetic energy to potential energy. By the use of the roller coaster, we can show that the total mechanical energy in the system is a constant.

As such we have that;

mgh = 1/2mv^2

m = mass of the object

g = acceleration due to gravity

h = height

v = velocity

Hence;

gh= 1/2v^2

v = √2gh

v = √ 2 * 9.8 * 49

v = 31 m/s

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Answer:

1A

Explanation:

=> Current (I) = q/t

Here charge (q) = 10C and time = 10sec

=> I = 10/10 = 1A

equal in magnitude but opposite in direction

Both of them are in big trouble !  I personally would write up each one of them for => no current valid driver's license, => no current vehicle insurance, => no current vehicle emission inspection sticker, => no visible current vehicle, => failure to signal turns, and => speeding in a school zone, since 19.4 m/s is equivalent to 43.4 mph.  

Answer:

isobaric

Explanation:

According to the attached figure, the process would be isobaric as the gas would be expanded at the time when it maintains the constant pressure

The ideal gas law derives

PV = nRT

As the flame is there at the bottom side so atoms or molecules of gas would get heat up slowly also their temperature would be increased. Thus in the case when piston would not be permitted to move so the pressure would rise having maintaining the constant volume and after this the process could be isochronic but in the given case, the piston would be permitted to move slowly upward so that there is a rise in the volume. However, the gas particles would be adjust themselves in order to maintain constant pressure so the given process represent isobaric  

To determine the image characteristics using the 3 principal rays and SALTS (Size, Attitude, Location, Type), we'll consider both lenses and mirrors separately. Here's how you can analyze each case:

Lenses:

Place an object at different distances to the left of a lens with a focal length (f).

a) Object placed beyond 2f:

In this case, the object is placed far beyond twice the focal length of the lens.

Principal ray 1: A ray parallel to the principal axis will pass through the focal point on the opposite side.

Principal ray 2: A ray passing through the optical center will continue in a straight line without any deviation.

Principal ray 3: A ray passing through the focal point on the object side will emerge parallel to the principal axis.

The image will be formed on the opposite side of the lens, between the focal point and twice the focal length.

SALTS:

Size: The image will be smaller than the object.

Attitude: The image will be inverted.

Location: The image will be located between the focal point and twice the focal length.

Type: The image will be real.

b) Object placed at 2f:

In this case, the object is placed at twice the focal length of the lens.

Principal ray 1: A ray parallel to the principal axis will pass through the focal point on the opposite side.

Principal ray 2: A ray passing through the optical center will continue in a straight line without any deviation.

Principal ray 3: A ray passing through the focal point on the object side will emerge parallel to the principal axis.

The image will be formed on the opposite side of the lens at twice the focal length.

SALTS:

Size: The image will be the same size as the object.

Attitude: The image will be inverted.

Location: The image will be located at twice the focal length.

Type: The image will be real.

c) Object placed between f and 2f:

In this case, the object is placed between the focal point and twice the focal length of the lens.

In this case, the object is placed far beyond twice the focal length of the mirror.

Principal ray 1: A ray parallel to the principal axis will reflect through the focal point on the same side.

Principal ray 2: A ray passing through the focal point on the object side will reflect parallel to the principal axis.

Principal ray 3: A ray passing through the center of curvature will reflect back along the same path.

The image will be formed on the opposite side of the mirror, between the focal point and twice the focal length.

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The diameter of the smaller end of the pipe is approximately 0.78 meters.

To determine the diameter of the smaller end of the pipe, we can use the principle of conservation of mass. According to this principle, the mass flow rate of water should remain constant throughout the pipe.

The mass flow rate is given by the equation:
Mass flow rate = density of water * cross-sectional area * velocity

Since the density of the water remains constant, we can write:
Cross-sectional area1 * velocity1 = Cross-sectional area2 * velocity2

Given that the velocity1 is 13 m/s, the diameter1 is 1.2 m, and the velocity2 is 30 m/s, we can solve for the diameter2 using the equation:
(pi * (diameter1/2)^2) * velocity1 = (pi * (diameter2/2)^2) * velocity2

Simplifying the equation:
(1.2/2)^2 * 13 = (diameter2/2)^2 * 30

Calculating the equation:
(0.6)^2 * 13 = (diameter2/2)^2 * 30

0.36 * 13 = (diameter2/2)^2 * 30

4.68 = (diameter2/2)^2 * 30

Dividing both sides by 30:
0.156 = (diameter2/2)^2

Taking the square root of both sides:
0.39 = diameter2/2

Multiplying both sides by 2:
0.78 = diameter2

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Answer:

a.\(F=7.83\times 10^{-51} N\)

b.Attractive

Explanation:

We are given that

\(F=\frac{GM_1M_2}{R^2}\)

\(G=6.67\times 10^{-11} Nm^2/kg^2\)

Mass of an electron,\(M_1=9.11\times 10^{-31} kg\)

Mass of proton,\(M_2=1836\times 9.11\times 10^{-31} kg\)

Distance between electron and proton,R=\(3.602nm=3.602\times 10^{-9} m\)

\(1nm=10^{-9} m\)

a.Substitute the values then  we get

\(F=\frac{6.67\times 10^{-11}\times 9.11\times 10^{-31}\times 1836\times 9.11\times 10^{-31}}{(3.602\times 10^{-9})^2}\)

\(F=7.83\times 10^{-51} N\)

b.We know that like charges repel to each other and unlike charges attract to each other.

Proton and electron are unlike charges therefore, the force between proton and electron is attractive.

Answer:

Explanation:

b)  Fg = W = mg = (104 kg)(9.8 m/s²) = 1019.2 N  (weight of the box)

Ff = (coeff. friction)(N)

N = normal force = Fg = 1019.2 N

c)  coeff. friction = Ff/N = 45 N/1019.2 N = 0.044

d)  Fnet = 245 N - 45 N = 200 N

It's hard to draw a free body diagram with the available symbols, but here's a try at it.  Just add a box between the arrows!

                            W

                            ↓

     Ff  →                                 ←  Fapplied

                             ↑

                             N